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Let $X_t$ be a random process such that \begin{eqnarray} X_1 &=& 0\\ X_t &=& X_{t-1} + \left\{\begin{array}{ll} A_t, & X_{t-1} \geq 0\\ B_t, & X_{t-1} < 0 \end{array}\right. \end{eqnarray} where $A_t$ and $B_t$ are i.i.d. random variables with the properties \begin{eqnarray} E[A_t] &=& {-}\alpha ,& \text{Var}[A_t] &= \sigma^2\\ E[B_t] &=& \alpha ,& \text{Var}[B_t] &= \sigma^2. \end{eqnarray} Question: I am looking for techniques to establish an upper bound on $E[|X_t|]$ or $E[X_t^2]$ which holds for all $t\in\mathbb{N}$ in terms of $\alpha$ and $\sigma$.

I can write out $E[X_t^2]$ as follows \begin{align} E[X_t^2] &= E\left[(X_{t-1}+1_{X_{t-1}\geq 0}A_t+1_{X_{t-1}<0}B_t)^2\right]\\ &= E[X_{t-1}^2] + E\left[(1_{X_{t-1}\geq 0}A_t+1_{X_{t-1}<0}B_t)^2\right] + 2 E\left[X_{t-1}(1_{X_{t-1}\geq 0}A_t+1_{X_{t-1}<0}B_t)\right]\\ &= E[X_{t-1}^2] + \alpha^2 + \sigma^2 - 2 E[|X_{t-1}|]\alpha \end{align} where $1_{\text{condition}}$ is the indicator function and the independence of $X_{t-1}$ and $A_t,B_t$ is used in the last equality. For the case $\sigma>0$ and by assuming that $X_{t-1}$ and $X_t$ are identically distributed in the limit $t\rightarrow \infty$, we have $E[X_t^2] = E[X_{t-1}^2]$. Hence, \begin{align} \lim_{t\rightarrow \infty}E[|X_t|] = \frac{\alpha^2+\sigma^2}{2\alpha}. \end{align}

In the case $\sigma= 0$, we have $X_t = -\alpha$ for even $t$ and $X_t = 0$ for odd $t$. Thus, $E[|X_t|] \leq \alpha = \frac{\alpha^2+\sigma^2}{\alpha}$.

This leads me to the conjecture: $E[|X_t|] \leq \frac{\alpha^2+\sigma^2}{\alpha}$ for all t.

I am also interested in the rate of convergence to the limit $\frac{\alpha^2+\sigma^2}{2\alpha}$ if anything can be said about that?

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  • $\begingroup$ Maybe this article is helpful: The oscillating random walk $\endgroup$ Commented May 18, 2016 at 21:39
  • $\begingroup$ Still interested in a solution to this ? $\endgroup$ Commented Jun 3, 2019 at 2:45

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